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Question

# In the given figure, AB || CD. If ∠BAE = 100° and ∠ECD = 120° then x = ? Figure

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Solution

## Draw $EF\parallel AB\parallel CD$. Then, $\angle AEF+\angle CEF=x°$ Now, $EF\parallel AB$ and AE is the transversal. $\therefore \angle AEF+\angle BAE=180°\left[\text{Sum of consecutive interior angles is supplementary}\right]\phantom{\rule{0ex}{0ex}}⇒\angle AEF+100°=180°\phantom{\rule{0ex}{0ex}}⇒\angle AEF=80°$ Again, $EF\parallel CD$ and CE is the transversal. $\angle CEF+\angle ECD=180°\left[\text{Sum of consecutive interior angles is supplementary}\right]\phantom{\rule{0ex}{0ex}}⇒\angle CEF+120°=180°\phantom{\rule{0ex}{0ex}}⇒\angle CEF=60°$ Therefore, $x°=\angle AEF+\angle CEF\phantom{\rule{0ex}{0ex}}=\left(80+60\right)°\phantom{\rule{0ex}{0ex}}=140°\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},x=140$

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